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Proof step 2 from proof mon-lem (previous step)

⌜n⌝ < # σσ, mon(σσ), m ≤ n ⊦ σσ[m] ⊑ σσ[n]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
⌜n⌝ < # σσ, mon(σσ), m ≤ n ⊦ ∃ σ1. σσ[m] + σ1 = σσ[n]

Rule argument

Additional information

Used simplifier rules: prefix-def: ⊦ σ ⊑ σ0 ↔ (∃ σ1. σ + σ1 = σ0);